Two concentric circles are radii 5cm and 3cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
Answers
Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.
It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.
AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)
therefore, ΔODB is a right angled triangle
where, OD²+BD²=OB²
(3)²+BD²=(5)²
9+BD²=25
BD²=25-9
BD²=16
BD=4cm
Since AD=BD=4cm
therefore, AB=AD+BD
AB=4+4
AB=8cm
Answer:
8 cm
Step-by-step explanation:
the radius of the smaller circle acts a a height for a right angle triangle and the radius of the larger circle acts as a hypotenuse if we join the 3 cm radius and 5 cm radius with the chord.
acc to pythagoras theorem, a^2 + b^2 = c^2
let x be the half of the chord we are finding (radius of 3 cm acts as a perpendicular to the chord)
so, x^2 + 3^2 = 5^2
x^2 = 25-9
x^2 = 16
x = + or - 4 but since a chord cannot be negative, x= 4 cm
now this is half of the chord we need, so, chord = 2 x 4 = 8 cm